How many members of the set S={x:1≤x≤100} are of the form 4n-1. I guess the answer is 25, but is there a way to prove it or show it?

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How many members of the set S={x:1≤x≤100} are of the form 4n-1. I guess the answer is 25, but is there a way to prove it or show it?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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n is an integer
Take the series of such numbers to be S. The nth term of that series will be given by 4n-1. Putting n=1,n=2 etc..you find that the series is: 3,7,11....... This is an arithmetic progression. 4n-1<100 4n<101 n<25.25 Therefore number of terms of the series would be 25.
i was trying some sigma sum notation but i couldn't get it to work Good answer @Mani_Jha

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{3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 71, 75, 79, 83, 87, 91, 95, 99} Here they are and they are 25. For proof, one proof above is right.

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